The reaction of 52 g of NH 3 with excess oxygen yields 45 g of NO according to the balanced equation below. What is the percent yield? 4 NH3 + 5 O2 → 4 NO + 6 H2O a. 83% b. 100% c. 18% d. 25% e. 49%

Oprion e. 49 %

Explanation:

Reaction is: 2NH₃ + O₂ → 2NO + 3H₂

Ratio is 2:2. 2 moles of ammonia can produce 2 moles of NO

We convert the mass to moles: 52 g. 1mol / 17g = 3.06 moles

Therefore, 3.06 moles of NO are produced, by the 100 % yield reaction.

3.06 moles. 30 g/1mol = 91.8 g

We have produced 45 g of NO, so to determine the percent yield:

(Yield produced / Theoretical yield). 100 =

(45 g / 91.8 g). 100 = 49%